# Probability current in 3-d

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.41.

We’ve seen the concept of the probability current in one dimension. A generalization to 3-d gives the vector

$\displaystyle \mathbf{J}=\frac{i\hbar}{2m}(\Psi\nabla\Psi^*-\Psi^*\nabla\Psi) \ \ \ \ \ (1)$

We can use the product rule from vector calculus to get

 $\displaystyle \nabla\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}(|\nabla\Psi|^{2}+\Psi\nabla^{2}\Psi^*-|\nabla\Psi|^{2}-\Psi^*\nabla^{2}\Psi)\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}(\Psi\nabla^{2}\Psi^*-\Psi^*\nabla^{2}\Psi)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Psi\left(-\frac{\partial\Psi^*}{\partial t}+\frac{i}{\hbar}\Psi^*V\right)-\Psi^*\left(\frac{\partial\Psi}{\partial t}+\frac{i}{\hbar}\Psi V\right)\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial}{\partial t}|\Psi|^{2} \ \ \ \ \ (5)$

where in the third line we used the Schrodinger equation rearranged to:

$\displaystyle \frac{i\hbar}{2m}\nabla^{2}\Psi=\frac{\partial\Psi}{\partial t}+\frac{i}{\hbar}\Psi V \ \ \ \ \ (6)$

Again generalizing the 1-d case, if we integrate the outward component of ${\mathbf{J}}$ over the surface area of a volume, this gives us the net rate at which probability ‘flows’ across the surface, and should, therefore, be equal to the rate at which probability changes within the volume. From the divergence theorem, we know that

 $\displaystyle \int_{A}\mathbf{J}\cdot d\mathbf{a}$ $\displaystyle =$ $\displaystyle \int_{V}\nabla\cdot\mathbf{J}d^{3}\mathbf{r}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{d}{dt}\int_{V}\left|\Psi\right|^{2}d^{3}\mathbf{r} \ \ \ \ \ (8)$

Since ${\left|\Psi\right|^{2}}$ is the probability density, this equation expresses exactly this idea.

As an example, we can look at the ${\psi_{211}}$ state of the hydrogen atom. We worked out this function earlier, so we quote the result here:

 $\displaystyle \psi_{211}$ $\displaystyle =$ $\displaystyle -\frac{1}{8a^{2}\sqrt{\pi a}}re^{-r/2a}\sin\theta e^{i\phi} \ \ \ \ \ (9)$

The gradient in spherical coordinates is

$\displaystyle \nabla=\hat{\mathbf{r}}\frac{\partial}{\partial r}+\frac{1}{r}\hat{\mathbf{\theta}}\frac{\partial}{\partial\theta}+\frac{1}{r\sin\theta}\hat{\mathbf{\phi}}\frac{\partial}{\partial\phi} \ \ \ \ \ (10)$

so we can calculate ${\nabla\psi_{211}}$:

$\displaystyle \nabla\psi_{211}=\frac{-1}{8a^{2}\sqrt{\pi a}}e^{-r/2a}e^{i\phi}\left[\left(1-\frac{r}{2a}\right)\sin\theta\hat{\mathbf{r}}+\cos\theta\hat{\mathbf{\theta}}+i\hat{\mathbf{\phi}}\right] \ \ \ \ \ (11)$

We can now calculate ${\mathbf{J}}$ from 1. The contributions along the ${\hat{\mathbf{r}}}$ and ${\hat{\mathbf{\theta}}}$ directions cancel out, so we are left with

 $\displaystyle \mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2m}\left[-\frac{1}{8a^{2}\sqrt{\pi a}}re^{-r/2a}\sin\theta e^{i\phi}\frac{i}{8a^{2}\sqrt{\pi a}}e^{-r/2a}e^{-i\phi}-\frac{1}{8a^{2}\sqrt{\pi a}}re^{-r/2a}\sin\theta e^{-i\phi}\frac{i}{8a^{2}\sqrt{\pi a}}e^{-r/2a}e^{i\phi}\right]\hat{\mathbf{\phi}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{64\pi ma^{5}}re^{-r/a}\sin\theta\hat{\mathbf{\phi}} \ \ \ \ \ (13)$

From 1, we see that the dimensions of ${\mathbf{J}}$ are those of velocity, since ${\hbar}$ has dimensions of ${\mbox{energy}\times\mbox{time}=\mbox{mass}\times\mbox{length}^{2}\times\mbox{time}^{-1}}$ so ${\frac{\hbar}{m}\nabla}$ has dimensions of ${\mbox{length}\times\mbox{time}^{-1}}$. The quantity ${m\mathbf{J}}$ therefore has dimensions of momentum, and can be interpreted as the flow of mass. This makes sense, since the probability density is the probability density of finding the particle within a given infintesimal volume, and the particle consists of its mass.

From ${\mathbf{L}=\mathbf{r}\times\mathbf{p}}$, we can therefore write the (orbital, ignoring spin) angular momentum as

$\displaystyle \mathbf{L}=\int_{V}\mathbf{r}\times\left(m\mathbf{J}\right)d^{3}\mathbf{r} \ \ \ \ \ (14)$

Since ${\mathbf{J}}$ is parallel to ${\hat{\mathbf{\phi}}}$ and ${\hat{\mathbf{r}}\times\hat{\mathbf{\phi}}=-\hat{\mathbf{\theta}}}$, we have for the hydrogen atom above:

$\displaystyle \mathbf{r}\times\mathbf{J}=\frac{-\hbar}{64\pi ma^{5}}e^{-r/a}r^{2}\sin\theta\hat{\mathbf{\theta}} \ \ \ \ \ (15)$

If we want to find ${L_{z}}$, we need the component of ${\mathbf{r}\times\mathbf{J}}$ along the unit vector ${\hat{\mathbf{k}}}$. The component of ${\hat{\mathbf{\theta}}}$ along ${\hat{\mathbf{k}}}$ is ${-\sin\theta\hat{\mathbf{k}}}$ so to find ${L_{z}}$ we must do the integral

 $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle m\int|\mathbf{r}\times\mathbf{J}|\sin\theta d^{3}\mathbf{r}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{64\pi a^{5}}\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}r^{4}e^{-r/a}\sin^{3}\theta drd\theta d\phi\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hbar \ \ \ \ \ (18)$

where we used Maple to do the integral. This is correct for the state ${\psi_{211}}$ since for quantum number (not mass) ${m=+1}$ we would expect ${L_{z}=+\hbar}$.

## 5 thoughts on “Probability current in 3-d”

1. Jilan

Hi, I know it doesn’t change the answer, as it disappears, but I got a factor of sin theta is my r term on the line level with the last week of the calendar.